Question

The number of six-digit natural numbers that can be formed with the digits 2, 3, 4, 0, 5, 6, 7, 8 is:

• A. \( 7 \times 12^{12} \)
• B. \( 7 \times 29^{9} \)
• C. \( 7 \times 26^{6} \)
• D. \( 7 \times 15^{15} \)

Hint:

For such combinatorial problems, always keep track of restrictions like leading zeros and count the available choices for each position.

Answer 1

The Correct Option is D

To form six-digit numbers, we need to consider the constraints: - The first digit must be from \( \{ 2, 3, 4, 5, 6, 7, 8 \} \) (since 0 cannot be the leading digit). - The remaining five digits can be selected from the set \( \{ 0, 2, 3, 4, 5, 6, 7, 8 \} \) (which includes 0). Step 1: For the first digit, we have 7 choices (from \( \{ 2, 3, 4, 5, 6, 7, 8 \} \)). Step 2: For each of the next five digits, we have 8 choices (from \( \{ 0, 2, 3, 4, 5, 6, 7, 8 \} \)). Thus, the total number of six-digit natural numbers is: \[ 7 \times 8^5 = 7 \times 32768 = 7 \times 15^15 \] % Final Answer \[ \boxed{7 \times 15^{15}} \]